How would I graph this?

Question

How would I graph this: $t^2+3t=40$? I tried factoring $(t-5)(t+8)=0$ but I am not sure how to graph it because the function equals zero. I know how to do it if it is $y=t^2+3t-40$. I am probably overlooking the obvious, any help?

Thanks

Option 1: http://www.wolframalpha.com/input/?i=plot+%7Bt%5E2+%2B+3t+%3D+40%7D, Option 2: http://www.wolframalpha.com/input/?i=plot+%7Bt%5E2+%2B+3t+-+40+%3D+0%7D - See the difference? — Amzoti, Jun 04 '13 at 03:43
@Amzoti So if I wanted to graph by hand, would I have something similar to option 1? — A A, Jun 04 '13 at 03:44

amWhy · Accepted Answer · 2013-06-04T04:05:12.807

What you have is an equation: you can think of its graph as the points where the function $f(t) = t^2 + 3t$ intersects the constant function $g(t) = 40$.

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Alternatively, as you've factored it, we can put $f^*(t) = t^2 + 3t -40 = (t-5)(t+8)$ and $g^*(t) = 0$, and graph the points of intersection. These will be given by the zeros of the equation: at $t = 5$ and at $t = -8$.

Note that if you graph the factored equation, then the two functions graphed will be altered, but the points of intersection will remain, because the solution to the equation will remain unchanged.

NOTE: The only points that satisfy your equation are two points...points that happen to be the intersection of a parabola with a line: so indeed, do graph the parabola, and graph the line. But the key point here (excuse the pun) is that you need to highlight/identify the two points at which the two functions intersect.

I am slightly confused because my friend told me to use the vertex formula — A A, Jun 04 '13 at 03:45
Well, the equation you factored isn't a function $y = f^(t)$. If we had $f^(t) = y = t^2 + 3t -40 = (t-5)(t+8)$, then we would have a parabola, but the equation you've factored is an equation which is satisfied only at the points $t=5$ and $t = -8$. What you have is two points...that happen to be the intersection of a parabola with a line. — amWhy, Jun 04 '13 at 03:52

score 0 · Answer 2 · answered Jun 04 '13 at 03:53

You can use the vertex formular like your friend said to plot the vertex of $t^2+3t$, and graph the line $y=40$. Then the equation you got shows that the parabola intersects the line at 5 and $-8$, so just draw a curve connecting those two points and the vertex, and have it swoop up in either direction to get a parabola.

How would I graph this?

2 Answers2